# characterization of Principal ideal rings

I'm thinking about these statements and I would like to know if I am right.

1. A principal ideal ring $R$, by definition, is a ring whose ideals are principals, since $R$ is itself an ideal, $R=(x)$, for some $x\in R$.

2. If $R=(y)$ for some $y$, every ideal of $R$ is principal.

3. Using (1) and (2), the principal ideal rings are in this form $(x)$, for some element $x$ and vice and versa, every (x) is a principal ideal ring.

Am I wrong? I need some help.

Thanks a lot

• Every unital ring satisfies $R = (y)$ for some $y$. – Ink Jul 16 '13 at 4:26
• @Ink $R$ is not necessarily unital. – user42912 Jul 16 '13 at 4:36
• And your point is? – Ink Jul 16 '13 at 4:46
• @Ink I would like to know if what I said above is true, maybe I mistaken in some part. – user42912 Jul 16 '13 at 4:59
• There's an answer below that tells you why it's false. – Ink Jul 16 '13 at 5:00

Consider for example the ring of all polynomials $P(x,y)$ with real coefficients. Certainly the whole ring is a principal ideal, since it is generated by the polynomial $1$. But the ideal generated by $x$ and $y$ is non-principal.
• This is the ideal of all polynomials that vanish at $(0,0)$. Can you think of a generator? It would have to be a polynomial that is $0$ at $(0,0)$. Since $x$ is in the ideal, it would have to look like $kx$. But $y$ is not $kx$ times a polynomial. – André Nicolas Jul 16 '13 at 6:03
Property (2) is completely wrong because any commutative unital ring $R = (1)$ and so this would imply that any commutative unital $R$ is a principal ideal ring! This is certainly not true for example by considering $R = \Bbb{Z}[x]$ and $I = (2,x)$: If $I$ is principal then since $R$ is Noetherian the Krull Hauptidealsatz implies that $I$ has height one which is a contradiction, for
$$0 \subsetneqq (x) \subsetneqq (2,x).$$